Question

In a cyclic process, heat transfers are + 14.7 kJ, – 25.2 kJ, – 3.56 kJ and + 31.5 kJ. What is the net work for this cyclic process ?

Answer 1

Solution

For a cyclic process -

$\oint Q=\oint W$

So given data are in question

$\oint Q=$ = 147 –25.2 –3.56 + 31.5 =149.74 kJ

So we get 

∮ W = 149.74 kJ

Therefore the system does 149.74 kJ of work per cycle.